PHYS 222 SI Exam Review. What to do to prepare Review all clicker questions, but more importantly...
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Transcript of PHYS 222 SI Exam Review. What to do to prepare Review all clicker questions, but more importantly...
PHYS 222 SI Exam Review
What to do to prepare
β’ Review all clicker questions, but more importantly know WHY
β’ Review quizzes
β’ Make sure you know what all the equations do, and when to use them
β’ These equations are used exclusively in LRC circuits
β’ These equations are what let you find the major constants that do not change with time.
β’ Remember, capital letters are not time dependent.
π πππ =πβ2, πΌ πππ =
πΌβ2
β’ These equations are used to determine the root-mean square voltage and current in an LRC circuit
π=πΌ πππ ππ‘β’ These two equations assume that the current
in an LRC circuit is a maximum at t=0.β’ These equations tell the voltage and current
as a function of time.β’ To find remember to add the appropriate
phase constants to the cos term.
πΌ πππ£=2ππΌ
β’ Not mentioned in class really
AC Current section on the Equation Sheet
β’ All the capitalized letters do not change with time.
β’ For exampleβ¦none of these change with time:β V, I,
β’ To find the time dependent voltage and current, multiply by the appropriate time dependent equation.
How to determine v(t), i(t),
β’ The current phasor is parallel with the phasor.β’ leads , and lags β’ The voltage β I.E. if then
What happens to a circuit at resonant frequency?
β’ The voltage phasor is parallel with the current phasor (this does not usually happen)
β’ (recall that both do not depend on time)β’ The sum of the voltages across the inductor
and the capacitor equals 0.
Example #1
β’ At t=0, the current in the circuit is a maximum of 3 A.
β’ Thenβ¦
β’ Also note that without doing any math you know that and β Make sure you understand why.
Example #2
β’ Letβs say that in an LRC circuit,
β’ Also suppose that youβve calculated the phase angle to be, and
β’ Then
Example #3
β’ They tell you
β’ To calculate I, find Z, then use β’ To calculate , first find I, then use
β’ Finally, if you need time dependence, add the appropriate phase shift to the cos or sin.
π=ππ£β’ Equation relating power, current, and voltage
πππ£πππππ=12πΌππππ π=πΌ πππ π πππ πππ π=πΌ πππ
2 π
β’ Average power in an LRC AC circuit
π 2
π 1
=π 2
π 1
β’ Equations used to convert a voltage inside a transformer
π 1 πΌ 1=π 2 πΌ2β’ Current and voltage in a transformer
πΈ (π₯ , π‘ )=πΈπππ₯ cos (ππ₯βππ‘ ) οΏ½ΜοΏ½β’ The equations for electromagnetic radiation,
or in other words lightβ’ Note that the direction of propagation is +x.β’ Also note that
π=1
βπ0π0β’ Speed of light related to two constants
π’=12π0πΈ
2+ 12π0
π΅2=π0πΈ2=π΅2
π0
β’ Energy density
π=πΉπ΄
=1π΄ππππ‘
=ππ
=πΈπ΅π0π
β’ Equations relating the radiation pressure of an electromagnetic wave to the poynting vector and E and B.
πΊ=1π0
π¬Γπ©
β’ Poynting vector.β’ Note that the poynting vector is perpendicular
to both E and B
πΌ=πππ£πππππ=πΈπππ₯π΅πππ₯
2π0=πΈπππ₯2
2ππ0=12 β π0π0 πΈπππ₯
2 =12π0π πΈπππ₯
2
β’ The intensity of electromagnetic radiation, related to the E field.
β’ Equations relating the speed of light c, the wavelength of light , the frequency of light the angular frequency , and the wave number .
π=ππ£
β’ The speed of light in a medium of index of refraction .
β’ For example, in glass the speed of light is not equal to m/s, but instead itβs equal to m/s (
ππ=π ππ ππππππ‘πππβ’ The equations for reflection and refraction
πΌ=πΌπππ₯cos (π )2
β’ The equation for intensity of light through a diffraction grating.
β’
sin (πππππ‘ )=π2π1
β’ The equation used to find the critical angle between two interfaces.
β’ At angles equal to or greater than the critical angle, refracted rays stop going through the second medium. Instead they undergo total internal reflection.
β’ Sometimes light coming from one direction onto an interface doesnβt have a critical angle, but if the light goes the other direction, then the critical angle exists.
π‘ππππ=π2π1
β’ Equation used to find Brewsterβs Angle, also known as the Polarization Angle .
β’ This angle is where reflection stops happening.
1π +1
π β²=1π
β’ The master equation for both lenses AND mirrors
β’ ALWAYSβ¦s>0. (the distance from the real object to the vertex of the mirror, or from the real object to the lens)
β’ For MIRRORSβ¦β f>0 if the mirror is concave, if the mirror is convex, then
f<0. Also f=R/2.β sβ>0 if the image is on the same side as the outgoing rays
β’ For LENSESβ¦β f>0 if the lens is more converging, otherwise f<0 if the
lens is more diverging
π= π¦ β²
π¦=β π
β²
π β’ Magnification caused by a mirror or lens
π hπ π ππππππππππππ=π 2
β’ Focal point of a spherical mirror
1π=(πβ1 )( 1π 1β
1π 2
)β’ βLensmakerβs Equationβ
πππ
+πππ β²
=ππβπππ
β’ Look up
π= π¦ β²
π¦=β
πππ β²
ππ π
β’ Spherical fish bowls is the main application of this equation
π= πβ²
π
β’ M is the angular magnification of a telescope.β’ Recall that and
ππ πππ=π πβ’ This is the equation for two-source
interference, used to find where the bright fringes are.
β’ To find the dark fringes, replace m with (m+1/2)
ππ πππ=(π+ 12 )π
β’ Destructive interference.
π¦=ππ ππ
β’ Assuming small angles the equation on the previous slide gives this result, where R is the distance between the slits and the screen, d is the separation of the slits, m is an integer, and y is the height above the central interference maximum.
πΌ=πΌπππ₯cos ( π2 )2
β’ In single slit diffraction, you can use this equation to find the intensity as a function of the angle.
π=2π(π2βπ1 )π
=2π ππ ππππ
β’ For two sources of waves, this equation finds the phase angle between them, depending on the location of the point where you measure the interference of the two waves.
2 π‘=πππβ’ Thin film destructive reflection
2 π‘=(π+ 12 )ππ
β’ Thin film constructive reflectionβ’ Recall that is the wavelength in that medium
of index of refraction
ππ ππ π=π πβ’ Single-slit diffractionβ’ a is the width of the slit.β’ This equation gives diffraction minimaβ’ To get maxima, replace m with (m+1/2)
π½=2πππ ππ ππ
β’ This equation gives for diffraction, which can then be used to get the intensity of light at various points.
β’ Intensity difference caused by single-slit diffraction.
β’ is calculated from a different equation
β’ This equation combines the effects of two-slit interference and the diffraction caused by each of the slits independently.
π =πΞ π
=ππ
β’ Used to find the chromatic resolving power for a diffraction grating
π πππ1=1.22ππ·
β’ This is used to find the resolving power of a small circular hole of diameter D.
β’ is the location of the first minimum.
2ππ πππ=ππβ’ Used to find the location of maxima for
diffraction gratings
Answer: D
B, A
C, A
A,B
β’ B
A,C
B
β’ C,B
A